How to sum up unique values using case statements having certain conditions - sql

Question

I have a table that may have the same item but with different sizes, I would like to count those items with more than one size (e.g. marvel shirt with S, M sizes will count as "1"), but still be able to count how many S and M. I have 2 outcomes I would like to get. Please see below for more detail.
TABLE B
ITEM_NO ITEM
=========================
3130C MARVEL_SHIRT
1845C SPONGEBOB_BOXERS
A900C CK_COAT
A988C RIDER_JEANS
TABLE C
ITEM_NO SIZE
===============
3130C S
3130C M
1845C M
A900C L
A988C M -
I tried just counting it but it is incorrect since it counts how many distinct sizes are there
select (case substr(item_no, 5, 1)
when 'C' then 'clothes'
when 'T' then 'toys'
else 'misc'
end) inv,
count(item_no) total
,sum (case when C.size = 'S' then 1 else 0 end) AS small
,sum (case when C.size = 'M' then 1 else 0 end) AS med
,sum (case when C.size = 'L' then 1 else 0 end) AS large
,count (distinct C.size) AS multiple_sizes
from B left outer join C on B.item_no = C.item_no
group by substr(item_no, 5, 1);
Actual outcome (incorrect):
INV TOTAL Small Med Large Multiple_Sizes
==========================================================
CLOTHES 4 1 3 1 3
Desired/expected outcome:
INV TOTAL Small Med Large Multiple_Sizes
==========================================================
CLOTHES 4 1 3 1 1
Below is another possible desired outcome given this scenario: What if those with multiple sizes shouldn't be counted separately (i.e. Marvel shirt has multiple sizes, thus it won't count the S or M since it's already counted under Multiple_Sizes)?
INV TOTAL Small Med Large Multiple_Sizes
==========================================================
CLOTHES 4 0 2 1 1

Answer 1

You probably need to group twice (1) by item number (2) by item category:
SELECT SUBSTR(item_no, 5, 1) AS category
, COUNT(*) AS count_products
, SUM(count_small) AS small
, SUM(count_med) AS med
, SUM(count_large) AS large
, SUM(CASE WHEN count_small + count_med + count_large > 1 THEN 1 END) AS has_multiple
FROM (
SELECT b.ITEM_NO
, COUNT(CASE WHEN c.SIZE = 'S' THEN 1 END) AS count_small
, COUNT(CASE WHEN c.SIZE = 'M' THEN 1 END) AS count_med
, COUNT(CASE WHEN c.SIZE = 'L' THEN 1 END) AS count_large
FROM b
LEFT JOIN c ON b.item_no = c.item_no
GROUP BY b.ITEM_NO
) x
GROUP BY SUBSTR(item_no, 5, 1)
| category | count_products | small | med | large | has_multiple |
| C | 4 | 1 | 3 | 1 | 1 |
And the variation:
SELECT SUBSTR(item_no, 5, 1) AS category
, COUNT(*) AS count_products
, SUM(CASE WHEN count_small + count_med + count_large = 1 THEN count_small END) AS small
, SUM(CASE WHEN count_small + count_med + count_large = 1 THEN count_med END) AS med
, SUM(CASE WHEN count_small + count_med + count_large = 1 THEN count_large END) AS large
, SUM(CASE WHEN count_small + count_med + count_large > 1 THEN 1 END) AS has_multiple
FROM (
SELECT b.ITEM_NO
, COUNT(CASE WHEN c.SIZE = 'S' THEN 1 END) AS count_small
, COUNT(CASE WHEN c.SIZE = 'M' THEN 1 END) AS count_med
, COUNT(CASE WHEN c.SIZE = 'L' THEN 1 END) AS count_large
FROM b
LEFT JOIN c ON b.item_no = c.item_no
GROUP BY b.ITEM_NO
) x
GROUP BY SUBSTR(item_no, 5, 1)
| category | count_products | small | med | large | has_multiple |
| C | 4 | 0 | 2 | 1 | 1 |

Answer 2

--creando tabla
create table #temp (itemId int, size nvarchar(1))
--insertando valores
insert into #temp values (1,'S')
insert into #temp values (1,'M')
insert into #temp values (2,'M')
insert into #temp values (3,'L')
insert into #temp values (4,'M')
-- table of Different Item Codes
select
itemId
into #masDeUnItem
from
(select itemId,size from #temp group by itemId,size) t1
group by itemId
having count(1) > 1
-- Variable of Counting different Items
declare #itemsDistintos int
-- Providing Value to Variable
select #itemsDistintos = count(1) from
(
select * from #masDeUnItem
) t1
--Outcome 1
select count(distinct(itemId)) TOTAL
,
sum(case when size = 'S' then 1 else 0 end) SMALL
, sum(case when size = 'M' then 1 else 0 end) MEDIUM
, sum(case when size = 'L' then 1 else 0 end) LARGE
, #itemsDistintos as Multiple_Sizes
from #temp
--Outcome 2
select count(distinct(a.itemId)) TOTAL
,
sum(case when size = 'S' and b.itemId is null then 1 else 0 end) SMALL
, sum(case when size = 'M' and b.itemId is null then 1 else 0 end) MEDIUM
, sum(case when size = 'L' and b.itemId is null then 1 else 0 end) LARGE
, #itemsDistintos as Multiple_Sizes
from #temp a
left join #masDeUnItem b
on a.itemId = b.itemId